What is the third proportional to y/x and 1/x

Solution:

(i) Let the required third proportional be x

36, 18, x are in continued proportion

\Rightarrow 36 : 18 = 18 : x

\Rightarrow 36x = 18 x 18

\Rightarrow\ x=\frac{18\times18}{36}=9

Required proportional = 9

(ii) Let the required third proportional be x

5.25, 7, x are in continued proportion

\Rightarrow 5.25 : 7 = 7 : x

\Rightarrow 5x = 7 x 7

\Rightarrow\ x=\frac{7\times7}{5.25}

\Rightarrow x=\frac{49}{5.25}=\frac{28}{3}=9\frac{1}{3}

(iii) Let the required third proportional be x

Rs. 1.60 , 0.40 . Rs. x in continued proportion

\Rightarrow 1.60 x x = 0.40 x 0.40

\Rightarrow\ x=\frac{0.40\times0.40}{1.60}=0.1

(i) Consider x as the third proportional to 5, 10

5: 10 :: 10: x

It can be written as

5 × x = 10 × 10

x = (10 × 10)/ 5 = 20

Hence, the third proportional to 5, 10 is 20.

(ii) Consider x as the third proportional to 0.24, 0.6

0.24: 0.6 :: 0.6: x

It can be written as

0.24 × x = 0.6 × 0.6

x = (0.6 × 0.6)/ 0.24 = 1.5

Hence, the third proportional to 0.24, 0.6 is 1.5.

(iii) Consider x as the third proportional to Rs. 3 and Rs. 12

3: 12 :: 12: x

It can be written as

3 × x = 12 × 12

x = (12 × 12)/ 3 = 48

Hence, the third proportional to Rs. 3 and Rs. 12 is Rs. 48

(iv) Consider x as the third proportional to 5 ¼ and 7

5 ¼: 7 :: 7: x

It can be written as

21/4 × x = 7 × 7

x = (7 × 7 × 4)/ 21 = 28/3 = 9 1/3

Hence, the third proportional to 5 ¼ and 7 is 9 1/3.

If x, y and z are in continued proportion then y is called the mean proportional (or geometric mean) of x and z.

If y is the mean proportional of x and z, y^2 = xz, i.e., y = +\(\sqrt{xz}\).

For example, the mean proportion of 4 and 16 = +\(\sqrt{4 × 16}\)  = +\(\sqrt{64}\) = 8

If x, y and z are in continued proportion then z is called the third proportional.

For example, the third proportional of 4, 8 is 16.

Solved examples on understanding mean and third proportional

1. Find the third proportional to 2.5 g and 3.5 g.

Solution:

Therefore, 2.5, 3.5 and x are in continuous proportion.

 \(\frac{2.5}{3.5}\) = \(\frac{3.5}{x}\)

⟹ 2.5x = 3.5 × 3.5

⟹ x = \(\frac{3.5 × 3.5}{2.5}\)

⟹ x = 4.9 g

2. Find the mean proportional of 3 and 27.

Solution:

The mean proportional of 3 and 27 = +\(\sqrt{3 × 27}\) = +\(\sqrt{81}\) = 9.

3. Find the mean between 6 and 0.54.

Solution:

The mean proportional of 6 and 0.54 = +\(\sqrt{6 × 0.54}\) = +\(\sqrt{3.24}\) = 1.8

4. If two extreme terms of three continued proportional numbers be pqr, \(\frac{pr}{q}\); what is the mean proportional?

Solution:

Let the middle term be x

Therefore, \(\frac{pqr}{x}\) = \(\frac{x}{\frac{pr}{q}}\)

⟹ x\(^{2}\) = pqr × \(\frac{pr}{q}\) = p\(^{2}\)r\(^{2}\)

⟹ x = \(\sqrt{p^{2}r^{2}}\) = pr

Therefore, the mean proportional is pr.

5. Find the third proportional of 36 and 12.

Solution:

If x is the third proportional then 36, 12 and x are continued proportion.

Therefore, \(\frac{36}{12}\) = \(\frac{12}{x}\)

⟹ 36x = 12 × 12

⟹ 36x = 144

⟹ x = \(\frac{144}{36}\)

⟹ x = 4.

6. Find the mean between 7\(\frac{1}{5}\)and 125.

Solution:

The mean proportional of 7\(\frac{1}{5}\)and 125 = +\(\sqrt{\frac{36}{5}\times 125} = +\sqrt{36\times 25}\) = 30

7. If a ≠ b and the duplicate proportion of a + c and b + c is a : b then prove that the mean proportional of a and b is c.

Solution:

The duplicate proportional of (a + c) and (b + c) is (a + c)^2 : (b + c)^2.

Therefore, \(\frac{(a + c)^{2}}{(b + c)^{2}} = \frac{a}{b}\)

⟹ b(a + c)\(^{2}\) = a(b + c)\(^{2}\)

⟹ b (a\(^{2}\) + c\(^{2}\) + 2ac) = a(b\(^{2}\) + c\(^{2}\) + 2bc)

⟹ b (a\(^{2}\) + c\(^{2}\)) = a(b\(^{2}\) + c\(^{2}\))

⟹ ba\(^{2}\) + bc\(^{2}\) = ab\(^{2}\) + ac\(^{2}\)

⟹ ba\(^{2}\) - ab\(^{2}\) = ac\(^{2}\) - bc\(^{2}\)

⟹ ab(a - b) = c\(^{2}\)(a - b)

⟹ ab = c\(^{2}\), [Since, a ≠ b, cancelling a - b]

Therefore, c is mean proportional of a and b.

8. Find the third proportional of 2x^2, 3xy

Solution:

Let the third proportional be k

Therefore, 2x^2, 3xy and k are in continued proportion

Therefore,

\frac{2x^{2}}{3xy} = \frac{3xy}{k}

⟹ 2x\(^{2}\)k = 9x\(^{2}\)y\(^{2}\)

⟹ 2k = 9y\(^{2}\)

⟹ k = \(\frac{9y^{2}}{2}\)

Therefore, the third proportional is \(\frac{9y^{2}}{2}\).

● Ratio and proportion

10th Grade Math

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